Closed rectangle-of-influence drawings for irreducible triangulations

  • Authors:
  • Sadish Sadasivam;Huaming Zhang

  • Affiliations:
  • Computer Science Department, University of Alabama in Huntsville, Huntsville, AL;Computer Science Department, University of Alabama in Huntsville, Huntsville, AL

  • Venue:
  • TAMC'10 Proceedings of the 7th annual conference on Theory and Applications of Models of Computation
  • Year:
  • 2010

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Abstract

A closed rectangle-of-influence (RI for short) drawing is a straight-line grid drawing in which there is no other vertex inside or on the boundary of the axis parallel rectangle defined by the two end vertices of any edge Biedl et al [2] showed that a plane graph G has a closed RI drawing, if and only if it has no filled 3-cycle (a cycle of 3 vertices such that there is a vertex in the proper interior) They also showed that such a graph G has a closed RI drawing in an (n−1) ×(n−1) grid, where n is the number of vertices in G They raised an open question on whether this grid size bound can be improved [2] Without loss of generality, we investigate maximal plane graphs admitting closed RI drawings in this paper They are plane graphs with a quadrangular exterior face, triangular interior faces and no filled 3-cycles, known as irreducible triangulations [7] In this paper, we present a linear time algorithm that computes closed RI drawings for irreducible triangulations Given an arbitrary irreducible triangulation G with n vertices, our algorithm produces a closed RI drawing with size at most (n−3) ×(n−3); and for a random irreducible triangulation, the expected grid size of the drawing is $({22n \over 27}+O(\sqrt{n})) \times ({22n \over 27}+O(\sqrt{n}))$ We then prove that for arbitrary n≥4, there is an n-vertex irreducible triangulation, such that any of its closed RI drawing requires a grid of size (n−3) ×(n−3) Thus the grid size of the drawing produced by our algorithm is tight This lower bound also answers the open question posed in [2] negatively.